Quadratic Functions and Equations

-Objective

t Graph quadratic functions.

| Solve quadratic ■ equations.

NewVocabularyparabolaaxis of symmetryvertexquadratic equation completing the square Quadratic Formula

1 Graph Quadratic Functions The graph of a quadratic function is called a parabola. To graph a quadratic function, graph ordered pairs that satisfy the function.

Example t Graph a Quadratic Function Using a TableGraph f(x) = 2x2 — 8x + 4 by making a table of values.

PTTm Choose integer values for x and evaluate the function for each value.

X f(x) = 2x2 - 8 x + 4 tyt) (x, f(x))

0 f(0) = 2(0)2 - 8(0) + 4 4 (0,4)

1 W — 2(1 )2 — 8(1) + 4 - 2 (1 .-2 )2 f(2) = 2(2)2 - 8(2) + 4 - 4 (2, - 4 )3 f(3) = 2(3)2 - 8(3) + 4 - 2 (3, - 2 )4 f(4) = 2(4)2 - 8(4) + 4 4 (4,4)

CTffTW Graph the resulting coordinate pairs, and connect the points with a smooth curve.

The axis of symmetry is a line that divides the parabola into two halves that are reflections of each other. Notice that, because the parabola is symmetric about the axis of symmetry, points B and C are 4 units from the x-coordinate of the vertex, and they have the same y-coordinate.

The axis of symmetry intersects a parabola at a point called the vertex. The vertex of the graph at the right is A(3, —3).

axis o f sym m etry

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KeyConcept Graph of a Quadratic Function

Consider the graph of y= ax2 + bx+ c, where a + 0.

• The y-intercept is a(0)2 + 6(0) + cor c.

• The equation of the axis of symmetry is

2 a• The x-coordinate of the vertex is

2 a

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StudyTipGraphing Quadratic Functions You can always use a table of values to generate more points for the graph of a quadratic function.

s.t-imof Axis of Symmetry, ^intercept, and VertexUse the axis of symmetry, y-intercept, and vertex to graph/(.r) = x2 + 2x — 3.

RTBTI Determine a, b, and c.

f(x) = ax2 + bx + c

| | | ~ a = 1 ,6 = 2, and c — — 3

f(x) = lx 2 + 2x — 3

Use a and b to find the equation of the axis of symmetry.

x = —2a

- T H T ” - 1

Equation of the axis of symmetry

a = 1 and b = 2

ETEfflfl Find the coordinates of the vertex.

Because the equation of the axis of symmetry is x = — 1, the x-coordinate of the vertex is —1.

f(x) = x2 + 2x — 3 / ( - l ) = ( - 1 )2 + 2(—1) - 3 / ( - l ) = - 4

The vertex is at (—1, —4).

Original equation

Evaluate W f o r x = —1. Simplify.

[ jJ S H Find the y-intercept and its reflection.

Because c = —3, the coordinates of the y-intercept are (0, —3). The axis of symmetry is x = —1, so the reflection of the y-intercept is (—2, —3).

rT T fn Graph the axis of symmetry, vertex, y-intercept and its reflection. Find and graph one or more additional points and their reflections. Then connect the points with a smooth curve.

y

i

0 X

V — z, —31-3)1 1

( - 1 , - AI I t ’

The y-coordinate of the vertex of a quadratic function is the maximum or minimum value of the function. These values represent the greatest or least possible value that the function can reach.

KeyConcept Maximum and Minimum Values

The graph of f(x) = axd

• opens up and has a

• opens down and has

fix)

+ bx + c, where a =/= 0

minimum value when a > 0, and

a maximum value when a < 0.

a > 0

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a < 0

fix)maximum

A -0

The y-coor minimum

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V xM X

minimum

dinate of the vertex is the The y-coor alue. maximum

i V

dinate of the vertex is the value.

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P10 | LESSON 0 -3 | Q u a d ra tic Functions and Equations

The domain of a quadratic function is all real numbers. The range will either be all real numbers less than or equal to the maximum value or all real numbers greater than or equal to the minimum value.

StudyTipCheck Your Answers Graphically You can check your answers in Example 3 by graphing W = - 3 x2 + 12x + 1 1 .

fix) = —3x2 + 12x + 11

From the graph, you can see that the maximum value of the function is y = 23, the domain is R, and the range is y < 23, for

vyeR_______________

B ^ S uS E B Maximum and Minimum ValuesConsider f(x ) = —3x2 + 12x + 11.

a. Determine whether the function has a maximum or minimum value.

For this function, a = —3. Because a is negative, the graph opens down and the function has a maximum value.

b. Find the maximum or minimum value of the function.

The maximum value of the function is the y-coordinate of the vertex. To find the coordinates of the vertex, first find the equation of the axis of symmetry.

hx = —— Equation of the axis of symmetry2 a

12 a = —3 and b = 1 22(—3)= 2 Simplify.

Because the equation of the axis of symmetry is x = 2, the x-coordinate of the vertex is 2. You can now find the j/-coordinate of the vertex, or maximum value of the function, by evaluating the original function for x = 2.

f(x) = — 3x2 + 12x + 11 Original function

/(2) = -3 (2 )2 + 12(2) + 11 x= 2

= - 1 2 + 24 + 11 Multiply.

= 23 Simplify.

Therefore,/(x) has a maximum value of 23 at x = 2.

c. State the domain and range of the function.

The domain of the function is all real numbers or R. The range of the function is all real numbers less than or equal to the maximum value 23 or y < 23.

2 Solve Q uadratic Equations A quadratic equation is a polynomial equation of degree 2. The standard form of a quadratic equation is ax2 + bx + c = 0, where a =f= 0. The factors of a

quadratic polynomial can be used to solve the related quadratic equation. Solving quadratic equations by factoring is an application of the Zero Product Property.

KeyConcept Zero Product Property

For any real numbers a and b, if ab = 0, then either a = 0, b = 0, or both a and b equal zero.

w m m m Solve by FactoringSolve x2 — 8.v + 12 = 0 by factoring.

x2 — 8x + 12 = 0 Original equation

(x — 2){x — 6) = 0 Factor,

x — 2 = 0 or x — 6 = 0 Zero Product Property

x = 2 x = 6 Simplify.

The solutions of the equation are 2 and 6.

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Another method for solving quadratic equations is to complete the square.

KeyConcept Complete the Square

To complete the square for any quadratic expression of the form x2 + bx, follow the steps below.

E T T iffn Find one half of £>, the coefficient of x.

Square the result in Step 1.

t iE f jJ H Add the result of Step 2 to x* + bx.

WatchOut!Completing the Square When completing the square, the coefficient of the x2 term must be 1.

Solve by Completing the SquareSolve x2 — ix + 1 = 0 by completing the square.

x 2 - 4x + 1 = 0 Original equation

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Rewrite so that the left side is of the form x2 + bx.x 2 - 4x + 4 = - 1 + 4 Because = 4, add 4 to each side.

(x - 2 )2 = 3 Write the left side as a perfect square.x - 2 = ± V 3 Take the square root of each side.

x = 2 ± V 3 Add 2 to each side.

x = 2 + V 3 or x = 2 — V 3 Write as two equations.

~ 3 .73 « 0 .27 Use a calculator.

The solutions of the equation are approximately 0.27 and 3.73.

Completing the square can be used to develop a general formula that can be used to solve any quadratic equation of the form ax2 + bx + c = 0, known as the Quadratic Formula.

KeyConcept Quadratic Formula

The solutions of a quadratic equation of the form ax2 + bx + c = 0, where a ^ 0, are given by the following formula.

- A ± Vft2 - 4ac - 2 a

, B * ---------------------------------------------------------------------------------------------------------------------------------------------------- _ - ~r------------------------------- - — - I-----------------------— - —

Solve by Using the Quadratic Formula

StudyTipDiscriminant The expression b2 - 4ac is called the discriminant and is used to determine the number and types of roots of a quadratic equation. For example, when the discriminant is 0, there are two rational roots.

Solve x2 — 4x + 15 = 0 by using the Quadratic Formula.

_ — b ± \ ]b 2 — 4acla

- ( - 4 ) ± V ( - 4 ) 2 - 4(1)(15)2(1)

_ 4 ± V l6 - 60 2

4 ± \ / —44 2

4 ± 2 i V l l 2

= 2 ± V IT i

The solutions are 2 4- V I l i and 2 — V I l i .

Quadratic Formula

Replace a with 1 ,6 with —4, and cw ith 15.

Multiply.

Simplify.

V —44 = V 4 v T T or 2/VTT

Simplify.

P12 | Lesson 0 -3 | Q u ad ra tic Functions and Equations

Exercises = Step-by-Step Solutions begin on page R29.

Graph each equation by making a table of values. (Example 1)

1. fix) = x2 + 5x + 6 2. fix) = x2 — x — 2

3. fix) = 2x2 + x — 3 4, fix) = 3x2 + 4x — 5

5. fix) = x2 — x — 6 6. fix) = —x2 — 3x — 1

7. BASEBALL A batter hits a baseball with an initial speed of 80 feet per second. If the initial height of the ball is 3.5 feet above the ground, the function d{t) = 80f — 16f2 + 3.5 models the ball's height above the ground in feet as a function of time in seconds. Graph the function using a table of values. (Example 1)

Use the axis of symmetry, y-intercept, and vertex to graph each function. (Example 2)

8. f i x ) == x2 + 3x + 2 9. f i x ) = x2 -- 9x + 8

10. fix)--= x2 — 2x + 1 11. f i x ) — x2 -- 6x - 16

12. II= 2x2 — 8x — 5 13. f i x ) = 3x2 + 12x - 4

14. HEALTH The normal systolic pressure P in millimeters of mercury (mm Hg) for a woman can be modeled by P(x) = O.Olx2 + 0.05x + 107, where x is age in years.(Example 2)

a. Find the axis of symmetry, y-intercept, and vertex for the graph of P(x).

b. Graph P(x) using the values you found in part a.

Determine whether each function has a maximum or minimum value. Then find the value of the maximum or minimum, and state the domain and range of the function. (Example 3)

15. y

-8 -4 0 4 if

Ii f V \ —

fix) = - 2 x 2 + 3 x - 5

16. in 1I

u4

- 8 - 4 4\

\ \\

fix) = 5 x 2 + 3x— 2

17. fix ) = - x 2 + 3x - 5

19. fix ) = 2x2 + 4x + 7

21. fix ) = —3x2 - 2x - 1

18. fix ) = x2 - 5x + 6

20. fix ) = 6x2 + 3x - 1

22. fix ) = —5x2 + lOx - 6

Solve each equation by factoring. (Example 4)

23. x2 - lOx + 21 = 0 24. p2 — 6p + 5 = 0

25. x2 - 3x - 28 = 0 26. 4w2 + 19w - 5 = 0

27. 4r2 — r = 5 28. ?2 + 6e - 16 = 0

Solve each equation by completing the square. (Example 5)

29. x2 + 8x — 20 = 0 30. 2a2 + llfl - 21 = 0

31. z2 - 2 z — 24 = 0 32. p2 - 3p - 88 = 0

33) t2 - 3t - 7 = 0 34. 3g 2 - 12g = - 4

Solve each equation by using the Quadratic Formula.(Example 6)

36. f2 — 6f + 13 = 035. m2 + 12m + 36 = 0

37. 6m2 + 7m — 3 = 0

39. 4x2 — 2x + 9 = 0

38, c2 - 5 c + 9 = 0

40. 3p2 + 4p = 8

41. PHOTOGRAPHY Jocelyn wants to frame a croppedphotograph that has an area of 20 square inches with a uniform width of matting between the photograph and the edge of the frame as shown.

a. Write an equation to model the situation if the length and width of the matting must be 8 inches by 10 inches, respectively, to fit in the frame.

b. Graph the related function.C. What is the width of the exposed part of the matting x?

Solve each equation.

42. x2 + 5x — 6 = 0

44. x2 — llx + 24 = 0

46. —x2 + 4x — 6 = 0

48. x2 — 4x + 7 = 0

43. a2 — 13a + 40 = 0

45. q2 - 12q + 36 = 0

47. 7x2 + 3 = 0

49. 2x2 + 6x —3 = 0

50. PETS A rectangular turtle pen is 6 feet long by 4 feet wide. The pen is enlarged by increasing the length and width by an equal amount in order to double its area. What are the dimensions of the new pen?

NUMBER THEORY Use a quadratic equation to find two real numbers that satisfy each situation, or show that no such numbers exist.

51. Their sum is —17 and their product is 72.

52. Their sum is 7 and their product is 14.

53. Their sum is —9 and their product is 24.

54. Their sum is 12 and their product is —28.

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